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A238880
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The sequence is a succession of pairs of nonnegative integers. The digits in each pair form a palindromic pattern. The sequence starts with a(1) = 0 and is always extended with the smallest available integer not yet present in the sequence.
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3
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0, 10, 1, 11, 2, 12, 3, 13, 4, 14, 5, 15, 6, 16, 7, 17, 8, 18, 9, 19, 20, 102, 21, 112, 22, 122, 23, 32, 24, 42, 25, 52, 26, 62, 27, 72, 28, 82, 29, 92, 30, 103, 31, 113, 33, 133, 34, 43, 35, 53, 36, 63, 37, 73, 38, 83, 39, 93, 40, 104, 41, 114, 44, 144, 45, 54, 46, 64, 47, 74, 48, 84, 49, 94, 50, 105, 51, 115, 55, 155, 56, 65, 57, 75, 58, 85, 59, 95, 60, 106, 61, 116, 66, 166, 67, 76, 68, 86, 69, 96, 70, 107, 71, 117, 77, 177, 78, 87, 79, 97, 80, 108
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OFFSET
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1,2
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COMMENTS
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This sequence is a permutation of the nonnegative integers.
See A245234 for the inverse permutation.
Due to the offset 1, this isn't a permutation of the nonnegative integers stricto sensu (but a bijection from positive integers to nonnegative integers), and the only fixed point appears to be a(18) = 18.
If the offset was chosen to be zero, the list of fixed points of this sequence would be 0, 19, 20, 85, 885, 1007, 1017, 1027, 4603, 5353, ...
By construction/definition, the terms a(2k-1) are always smaller than all subsequent terms a(m), m > 2k-1, k > 0. Therefore, when the "reversal" A004086(a(2k-1)) > a(2k-1), then a(2k) = A004086(a(2k-1)): this obviously yields a palindromic pair (a(2k-1), a(2k)), and any smaller a(2k) with this property would have to be a left-truncation of this term, which would have less digits and therefore be strictly smaller and have occurred earlier than a(2k-1).
Otherwise, when A004086(a(2k-1)) <= a(2k-1), in particular when a(2k-1) is a palindrome or divisible by 10 or when A000030(a(2k-1)) > A010879(a(2k-1)) (first digit > last digit), then a(m) = A004086(a(2k-1)) for some m < 2k-1, and in this case, a(2k) = A004086(a(2k-1)*10+1) (except for a(2) = 10 where 10*a(1) vanishes) is the smallest possible number that yields a palindromic pair: again, any smaller solution would have to be a left truncation of this, which all have occurred earlier. (End)
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LINKS
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FORMULA
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a(2k) = A004086(a(2k-1)) if this is greater than a(2k-1), else A004086(a(2k-1)*10+1), for all k > 0.
A010879(a(2k)) > 0 for all k > 1: a(2) is the only even-indexed term multiple of 10. (End)
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EXAMPLE
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(0,10), (1,11), (2,12), (3,13), (4,14), (5,15), ...: The palindromic pattern is clearly visible in each pair.
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MATHEMATICA
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(* first 10000 terms *) s = {0, 10}; t = 0*Range[20000]; t[[10]] = 1; Do[ j = Position[t, 0, 1, 1][[1, 1]]; AppendTo[s, j]; t[[j]] = 1; d = IntegerDigits[j]; h = 1; While[t[[h]] == 1 || (p = Join[d, IntegerDigits[h]]; p != Reverse[p]), h++]; AppendTo[s, h]; t[[h]] = 1; If[Mod[Length@s, 500] == 0, Print[Length@s, " ", {j, h}]], {4999}]; s (* Giovanni Resta, Mar 06 2014 *)
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PROG
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(Haskell)
import Data.List (delete)
a238880 n = a238880_list !! (n-1)
a238880_list = f [0..] where
f (u:us) = u : g us where
g vs = h vs where
h (w:ws) = if reverse ys == ys then w : f (delete w vs) else h ws
where ys = xs ++ show w
xs = show u
(PARI)
A238880_upto(N, U=[-1])={vector(N, n, U=setunion(U, [N=if(n%2, U[1]+1, N < n=fromdigits(Vecrev(digits(N))), n, n+10^logint(N+!N, 10)*10)]); while(#U>1&& U[1]+1==U[2], U=U[^1]); N)} \\ M. F. Hasler, Apr 22 2024
(Python)
except AttributeError: A238880.terms = T = []; A238880.least_unused = 0
except IndexError: T = A238880.terms # more terms needed
while len(T) < n:
if len(T) & 1 == 0: x = A238880.least_unused # even index
elif T[-1] >= (x := int(s := str(T[-1])[::-1])): x = int('1'+s)
T.append(x)
if x == A238880.least_unused: # also e.g. for a(20) = 19
while x+1 in T: x += 1
return T[n-1] # M. F. Hasler, Apr 19 2024, simplified Apr 22, 2024
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CROSSREFS
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Cf. A371113 (analog for triples instead of pairs).
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KEYWORD
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AUTHOR
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STATUS
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approved
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